# Aptitude - Numbers - Discussion

Discussion Forum : Numbers - General Questions (Q.No. 32)

32.

How many 3-digit numbers are completely divisible 6 ?

Answer: Option

Explanation:

3-digit number divisible by 6 are: 102, 108, 114,... , 996

This is an A.P. in which *a* = 102, *d* = 6 and *l* = 996

Let the number of terms be *n*. Then *t*_{n} = 996.

*a* + (*n* - 1)d = 996

102 + (*n* - 1) x 6 = 996

6 x (*n* - 1) = 894

(*n* - 1) = 149

*n* = 150

Number of terms = 150.

Discussion:

32 comments Page 1 of 4.
Dhannvanth said:
6 months ago

My idea is;

1+4+9=14.

1+5+0=6.

1+5+1=7.

1+6+6=13.

So, according to me, here 6 only can be purely divisible by 6.

1+4+9=14.

1+5+0=6.

1+5+1=7.

1+6+6=13.

So, according to me, here 6 only can be purely divisible by 6.

(1)

Anand said:
2 years ago

Let the highest 3 digit num divided by 6 is 996,

then 996/6= 166.

Let the highest 2 digit num divided by 6 is 96,

then 96/6= 16,

Now 166-16= 150.

then 996/6= 166.

Let the highest 2 digit num divided by 6 is 96,

then 96/6= 16,

Now 166-16= 150.

(1)

Sahul hameed said:
2 years ago

As per the Divisibility rule; If last digit is even it will divided by 2 and 3 also divided by 6. In this problem only 166 can sacrifice this method. So the answer is 166.

Amar said:
3 years ago

Lowest 3 digit number which is divisible by 3 is 102.

Highest 3 digit number which is divisible by 3 is 999.

So as per the formula a+(n-1)d=l.

The answer will be 300.

Highest 3 digit number which is divisible by 3 is 999.

So as per the formula a+(n-1)d=l.

The answer will be 300.

Mahesh said:
4 years ago

6 = 2 * 3 which are co-actors, then the number divisible by these two will also divisible by 6.

Tirtha said:
4 years ago

By taking the multiples of 2*3 for 6 we can operate the divisibility rule for this question.

As for 2, the ending digit should be 0, 2, 4, 6, 8 and for 3 sum of all the digit should be divisible by 3.

So, 150 is divisible by both 2 & 3. This is the correct answer.

As for 2, the ending digit should be 0, 2, 4, 6, 8 and for 3 sum of all the digit should be divisible by 3.

So, 150 is divisible by both 2 & 3. This is the correct answer.

Tushar said:
5 years ago

Step 1) Consider highest 3 digit number divisible by 3 which is 996, dividing it by 6 we get remainder as 166. Since we want only 3 digit numbers which are divisible by 3 we need to subtract the 2 digit numbers in here.

Step 2) Consider highest 2 digit number divisible by 3 which is 96,dividing it by 6 we get remainder as 16.That means there are 16 (2 digit numbers) divisible by 6.

Step 3) Now subtract these two corresponding remainders we get 166-16 = 150.

Step 2) Consider highest 2 digit number divisible by 3 which is 96,dividing it by 6 we get remainder as 16.That means there are 16 (2 digit numbers) divisible by 6.

Step 3) Now subtract these two corresponding remainders we get 166-16 = 150.

Nithya said:
5 years ago

Divisiablity shortcut for 6 = the number should divide be both 2 & 3.

a. 149 / 2 = no because unit digit is odd.

b. 150 /2 = yes because unit digit ends in 0, then check for 3 -> 150 / 3 = 1+5+0 = 6 = 6/3 = 2, so answer is B.

a. 149 / 2 = no because unit digit is odd.

b. 150 /2 = yes because unit digit ends in 0, then check for 3 -> 150 / 3 = 1+5+0 = 6 = 6/3 = 2, so answer is B.

Vatsal said:
5 years ago

Number is divided by 6 only when it is perfectly divided by 2 and 3 both.

So Only 150 is divided by both 2 and 3.

So Only 150 is divided by both 2 and 3.

Damini said:
6 years ago

Write numbers from sequence 149 to 387 like 149150159.

How many time does 1 occur?

How many time does 1 occur?

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